# Find exact value of tan when given cos

Given $$\cos30 = \frac{\sqrt3}{2}$$ use trigonometric identities to find the exact value of $$\tan\frac{\pi}{3}$$

I understand that $$\cos30 = \frac{\sqrt3}{2}$$ from the standard trig values chart and I know that $$\frac{\pi}{3}$$ is 60 degrees and I know the value of it from the same chart. I'm not understanding how to use identities to find the value.

• What is "The" fundamental trigonometric identity? Once you know the sine of the angle you can easily compute the tangent. – Chris Leary Jan 19 at 22:06
• I don't understand where the "$\tan$" part in the question title comes in based on what is written in the question text. Is something missing? – John Omielan Jan 19 at 22:07
• Given your title, are you rather asking, in the body of your post, to find the exact value of $\tan\left(\pi/3\right)$? – jordan_glen Jan 19 at 22:07
• Sorry missed writing $\tan$ of – dstarh Jan 19 at 22:08

Upto the sign, you can calculate the tan-value as $$\tan(x)=\frac{\sqrt{1-\cos^2(x)}}{\cos(x)}$$

Also note $$\cos(2x)=2\cos^2(x)-1$$ which allows you to calculate $$\cos(\frac{\pi}{3})$$ from $$\cos(\frac{\pi}{6})$$

• But the question is asking $\tan\frac{\pi}{3}$ and you have $\cos \frac{\pi}{6}$ – user289143 Jan 19 at 22:12
• But $x$ is not $\pi/6$, but rather $\pi/3$. – jordan_glen Jan 19 at 22:13

Now $$\tan\frac{\pi}{3}=\frac{\sin\frac{\pi}{3}}{\cos\frac{\pi}{3}}=\frac{\cos\frac{\pi}{6}}{\sin \frac{\pi}{6}}=\frac{\cos \frac{\pi}{6}}{\sqrt{1-\cos^2\frac{\pi}{6}}}=\frac{\frac{\sqrt{3}}{2}}{\sqrt{1-\frac{3}{4}}}=\frac{\sqrt{3}}{2}\cdot \frac{1}{\sqrt{\frac{1}{4}}}=\sqrt{3}$$

Just to use another elementary way, which shows all the passages:

$$\cos(60) = \cos(30+30) = \cos(30)\cos(30) - \sin(30)\sin(30)$$

The value of $$\cos(30)$$ you know it.

The value of $$\sin(30)$$ is derived from $$\cos^2 + \sin^2 = 1$$.

Again, once you found $$\cos(60)$$ you get the value of $$\sin(60)$$ for free.

Now you can find $$\tan(60)$$.

\begin{align} &\color{red}{\tan\left(\frac\pi3\right)}=\frac{\sin\left(\frac\pi3\right)}{\cos\left(\frac\pi3\right)}=\frac{\sqrt{1-\cos^2\left(\frac\pi3\right)}}{\cos\left(\frac\pi3\right)}=\frac{\sqrt{1-\left(2\color{blue}{\cos\left(\frac\pi6\right)}^2-1\right)^2}}{2\color{blue}{\cos\left(\frac\pi6\right)}^2-1}\\ &=\frac{\sqrt{1-\left(2\left(\color{blue}{\frac{\sqrt 3}2}\right)^2-1\right)^2}}{2\left(\color{blue}{\frac{\sqrt 3}2}\right)^2-1}=\frac{\sqrt{1-\left(2\frac34-1\right)^2}}{2\frac34-1}=\frac{\sqrt{1-\left(\frac12\right)^2}}{\frac12}\\ &=\frac{\sqrt{\frac34}}{\frac12}=\frac{\frac12\sqrt3}{\frac12}=\color{red}{\sqrt 3} \end{align}

$$\therefore~\tan\left(\frac\pi3\right)=\sqrt3$$

$$\sin(\pi/3) = \sqrt 3/2,$$ and $$\cos(\pi/3) = 1/2$$.

Therefore, $$\tan(\pi/3) = \dfrac{\sin\left(\frac \pi 3\right)}{\cos\left(\frac \pi 3\right)} = \dfrac {\dfrac{\sqrt 3}2}{\dfrac 12}=\sqrt 3$$

Assume $$x = \frac\pi6$$, thus $$\frac\pi3 = 2x$$

$$\tan\left(2x\right) = \frac{\sin\left(2x\right)}{\cos\left(2x\right)} = \frac{2\sin\left(x\right)\cos\left(x\right)}{2\cos^2\left(x\right)-1} = \frac{2\sin\left(x\right)\cos\left(x\right)}{2\cos^2\left(x\right)-1} = \frac{2\sqrt{\left(1-\cos^2\left(x\right)\right)}\cos\left(x\right)}{2\cos^2\left(x\right)-1}$$

Since, $$\cos\left(x\right)=\cos\left(\frac\pi6\right)=\frac{\sqrt 3}2$$, substituting that on above equation, you'd get: $$\tan\left(2x\right) = \frac{2\sqrt{\left(1-\cos^2\left(x\right)\right)}\cos\left(x\right)}{2\cos^2\left(x\right)-1} = \frac{2\sqrt{\left(1-\frac 34\right)}\left(\frac{\sqrt{3}}2\right)}{2\times \frac 34-1} = \sqrt{3}$$

$$\therefore~\tan\left(\frac\pi3\right)=\sqrt3$$

Trigonometric Identities used are: $$\sin\left(2x\right) = 2\sin\left(x\right)\cos\left(x\right)$$ $$\cos\left(2x\right) = 2\cos^2\left(x\right)-1$$ $$\cos^2\left(x\right)+\sin^2\left(x\right)=1$$